Thank you so much. I appreciate it... 🙏🏽

These are different reference frames from which you are arguing. The statement that B does not move with respect to A is from the point of view of a rotating reference frame. The statement that B moves with 2ωr is from the point of view of a translational reference frame that moves tangentially to A.

I use Blender

This is just a notation I use to make the difference between a result and a condition clear. "=" ultimately means "is the result of something" and "=!" means "is a condition that we presuppose".

Each point on the gear is ultimately a superposition of the movement of the centre of gravity (which is the same for all points considered) and the rotational movement. The gear, and therefore the rotation, moves with the centre of gravity and not with the circumferential speed. The linear speed distribution (shown in blue) is thus shifted to the right by the amount of the centre of gravity speed. In fact, if the gear itself did not rotate around its center of gravity, which means that a marked tooth on the gear would always be at the 12 o'clock position as the center of gravity moves on a circular path, then every point on the gear would actually move at the same speed, regardless of the radius. The speed at each point on the gear would be the speed of the center of gravity.

@@tec-science Isn't Vc (at the center of gravity of the planet gear) equal to Nc (angular velocity of carrier) times the distance between the center of carrier arm and the center of planet gear? In the same way, will the velocity not vary throughout the planet gear (I am talking only about the velocity due to rotation of carrier arm. The velocity indicated in green colour in the video)?

I think I understand your problem. But the linear velocity distribution of the planet gear has its center of gravity as its reference. The linear velocity distribution is related to this center of gravity. If this center of gravity now moves with the velocity vc (velocity of the carrier), then the entire linear velocity distribution is superimposed by this velocity vc. The entire linear velocity distribution is shifted by this amount. Therefore, the velocity vc is added to the entire velocity distribution regardless of the radius considered on the planet gear.

@@tec-science I understand your explanation. You have superimposed the velocity due to rotation of planet gear due to its rotation about its center of gravity (blue velocity vectors)and the velocity due to the velocity of the center of gravity itself. However, will not the velocity Vc be constant (green velocity vectors) throughout only if the gear was moving in a straight line? In this case, the planet gear is undergoing revolution about the carrier center, not translation in a straight line. So, the green velocity distribution should also linearly vary (instead of being constant, as shown in the green distribution). Please correct me if I am wrong.

kzfaq.info/get/bejne/g7dpeNql1ridl4k.html I think your mistake is that when you superimpose the movements where the planet gear is initially assumed not to rotate, you think that the planet gear is firmly locked to the carrier. However, with this assumption, the planet gear rotates once around itself with each revolution of the carrier (this false assumption leads to the coin rotation paradox). When it is said that the planet gear does not rotate while the carrier rotates, this means that, for example, a marked tooth on the gear that is at the 12 o'clock position is there for the entire rotation (see the animation in the link). There is no linear speed distribution, but the planet gear has a constant speed along its entire diameter (shown in green).

β refers to the rotation of the planet gear when the sun gear is stationary. Now, however, the movement of the sun gear is superimposed on this and turns the planet gear backwards, so that the planet gear now effectively covers the angle 𝝳.